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Algebra: The Notes 1.1 Sequences and Sums An arithmetic sequence is a sequence of numbers a1, a2, a3,…an such that a1 – ai -1 = d for all i. Given a1, d, and n, an and Sn (the sum of the arithmetic sequence through n) can be found. an a1 (n 1)d Sn (a1 an )n 2 A geometric sequence is a sequence of numbers a1, a2, a3,…,an such that ai r for all i. ai 1 Given a1 and r, an, Sn and S∞ (the sum of the infinite series) can be found. a n a1 r n 1 1 rn 1 r a S 1 , given that |r| < 1 1 r Sn Sigma notation denotes the sum of successive values of a function f in the manner b f (i) f (a) f (a 1) f (a 2) ... f (b 1) f (b) i a The following are useful equations for sigma notation The number of values being summed is equal to b-a+1 b c c(b a 1) ia n i i 1 n i 2 i 1 n(n 1) 2 n(n 1)( 2n 1) 6 n 2 (n 1) 2 i 4 i 1 n 3 1.2 Exponents and Logarithms Exponents and logarithms are related in that if a b c then log a c b . Useful log and exponent rules: a loga x x log a a x x a x a y a x y log a b log a c log a bc a x a y a x y log a b log a c log a (a x ) y a xy log a b n n log a b Change of base: log a b b c log c b log c a 1.3 Counting and the binomial theorem Counting rules: Sum rule: the number of elements in two disjoint groups is the sum of the elements in each group Product rule: if you’re counting the number of ordered pairs of things as (A,B), where there are |A| choices for the first part and |B| choices for the second part, the total number of ordered pairs is |A| x |B| Subtraction rule: If you are counting the number of elements in some set A, it may be easier to count Ā (everything not in A) A permutation is an ordered sequence of some or all elements in a set n! is the number of permutations of every element in the set, where n is the number of elements in the set A combination is an unordered subset of elements from a given set n n! or n C k (n k )! k! k The binomial theorem gives the expansion of ( x y ) n , where n is a positive integer n n ( x y ) n x n k y k k 0 k 1.4 Proof by mathematical induction This technique allows proof of some statement for all non-negative or positive integer values of a variable. Induction is always done on a particular variable (ex. n) and involves three steps Statement: for all positive or non-negative integers n, S(n) 1. Base case: prove S(1), the smallest possible value for n. 2. Inductive hypothesis: Assume for an arbitrary positive or non-negative integer k, that S(k) is true. 3. Inductive step: Using the assumption in the previous step, prove that S(k+1) is true. Bear in mind that if a statement can be proved through induction, it must have a variable on which to do induction, and that variable must hold for either all positive or all non-negative integer values 1.5 Complex numbers 1 is defined as i. A complex number can be written in the form z a bi , where a is known as the real part and bi is known as the imaginary part. The conjugate of a complex number has the same real part but the sign is switched on the imaginary part. Ex. 3 – 4i has a conjugate 3 + 4i The Cartesian form of a complex number ( z a bi ) can be rewritten in the modulusargument form z r (cos i sin ) , where r is known as the modulus and (cos i sin ) is the argument. z r (cos i sin ) can also be written z rcis or z re i The complex plane, also known as the Argand plane, is a way of representing complex numbers. The x-axis represents the real part and the y-axis represents the imaginary part. By plotting a Cartesian complex number, its modulus-argument form can be determined. 1.6 Operations on complex numbers Complex numbers can be added or subtracted by simply adding or subtracting the real and imaginary parts separately and then combining them into the final answer To multiply two complex numbers, they should be multiplied like any numbers with two different terms: (a bi)(c di) ac bci dai bdi 2 ac bd (bc ad )i To divide two complex numbers, they must be put into fraction form and then both numerator and denominator must be multiplied by the conjugate of the denominator: a bi c di ac bci adi bdi 2 ac bd (bc ad )i c di c di c 2 d 2i 2 c2 d 2 1.7 De Moivre’s theorem De Moivre’s theorem states that an exponent of a complex number in modulus-argument form can be rewritten thus: (r (cos i sin )) n r n (cos n i sin n ) To take the power of a complex number, convert it into m-a form (if necessary) and then employ De Moivre’s theorem. Simplify cis term and then, if necessary, convert back into Cartesian form To find the nth root of a complex number, set r n (cis ) n equal to the number and then solve, using De Moivre’s theorem 1.8 Conjugate roots of polynomial equations with real coefficients A polynomial equation with n roots can be represented as follows 0 ( x r )( x s)( x t )... where r, s, t… are roots of the equation. The second coefficient of the equation is the opposite of the sum of the roots. Each coefficient, starting with the second is sum of 1,2,3… until the last coefficient is the product of all the roots for equations with an even number of roots and the opposite of the product for equations with an odd number of roots. Algebra: The Problems 1.1 Sequences and Sums For a geometric sequence, a2 is 3 and a5 is 81. Find a1, r and the infinite sum of this series 1.2 Exponents and logarithms log 4 x log 16 6 log 16 3 . Find x. 1.3 Counting and the binomial theorem Find the coefficient on the x5 term in the expression (2x + 7)9 1.4 Proof by mathematical induction 1 0 Find a formula for 0 5 n 1.5 Complex numbers Rewrite 5 2 (cis ) as a Cartesian form complex number 4 1.6 Operations on complex numbers Divide 3 + 5i by 4 + 7i. 1.7 De Moivre’s theorem Find all solutions to the equation z 5 16 16 3 . 1.8 Conjugate roots of polynomial equations with real coefficients Let r, s, and t represent the roots of the equation x 3 7 x 2 11x 49 Without calculating r, s, or t find r 2 s 2 t 2 Algebra: The Solutions 1.1 Sequences and Sums a5 r3 a2 81 27 r 3 3 r 3 a n a1 r n 1 3 a1 r 1 3 a1 3 a1 1 Since |r|>1, this sequence cannot be summed to infinity 1.2 Exponents and logarithms log 16 6 log 16 3 log 16 2 log 2 .25 log 16 log 4 x .25 log 16 2 4 .25 x 2 1.3 Counting and the binomial theorem n n ( x y ) n x n k y k k 0 k x 2x Coefficient of x5 is 9 (2 x) 5 7 4 4 y7 126 32 x 5 2401 n9 9680832 x 5 nk 5 k4 1.4 Proof by mathematical induction 2 1 0 1 0 0 5 0 25 3 1 0 1 0 0 5 0 125 Prove for all non-negative integers that n 1 0 1 0 0 5 0 5 n 0 B.C. 1 0 1 LHS 0 5 0 1 0 1 RHS 0 0 5 0 0 1 0 1 I.H. Assume for all non-negative integers k that k 1 0 1 0 0 5 0 5 k I.S. Prove, given the above, that 1 0 0 5 k 1 1 0 k 1 0 5 0 1 k 0 5 5 1 0 1 0 k 0 5 0 5 k 1 0 1 0 according to the I.H. 0 5 0 5 1 0 0 5 k 1 1.5 Complex numbers 5 Length 5 2 4 5 Cartesian form 5 5i 1.6 Operations on Complex numbers 3 5i 4 7i 12 12i 35i 2 47 12i 4 7i 4 7i 65 16 49i 2 1.7 De Moivre’s theorem z 5 16 16 3 z 5 32cis 3 r (cis ) 32cis 3 5 r 32 5 r2 cis (5 ) cis 3 5 5 3 2k , k Z 2 k , k 0,1,2,3,4 5 5 3 5 7 9 , , , , 5 5 5 5 5 3 5 7 9 z 2cis ,2cis ,2cis ,2cis ,2cis 5 5 5 5 5 1.8 Conjugate roots of polynomial equations with real coefficients x 3 7 x 2 11x 49 r s t 49 rs st tr 11 rst 49 (r s t ) 2 r 2 s 2 t 2 2(rs st tr ) r 2 s 2 t 2 (r s t ) 2 2(rs st tr ) r 2 s 2 t 2 (7) 2 2(11) r 2 s 2 t 2 49 22 71